Initial Value Theorem MCQ Quiz - Objective Question with Answer for Initial Value Theorem - Download Free PDF

Initial value and final value theorems of z transform are defined only for causal signals, i.e. for signals which are 0 for n<0.

Since the given signal is defined for n<0, we cannot use the initial value theorem here.

Derivation:

The z-transform for a causal signal is defined as:

$X\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}x\left(n\right)z^{-n}$

Expanding the above series, we get:

X(z) = x(0) + x(1) z^-1 + x(2) z^-2 + ....

$\underset{z\to \mathrm{\infty }}{lim}X(z)=x(0)+x(1){\mathrm{\infty }}^{-1}+x(2){\mathrm{\infty }}^{-2}+...$

This can be written as:

$\underset{z\to \mathrm{\infty }}{lim}X(z)=x(0)$

Since x(0) is the initial value of the given signal, this is the initial value theorem of z-transform.

Since, We cannot apply the z transform, none of these is the correct answer. 

Initial value and final value theorems of z transform are defined only for causal signals, i.e. for signals which are 0 for n<0.

Since the given signal is defined for n<0, we cannot use the initial value theorem here.

Derivation:

The z-transform for a causal signal is defined as:

$X\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}x\left(n\right)z^{-n}$

Expanding the above series, we get:

X(z) = x(0) + x(1) z^-1 + x(2) z^-2 + ....

$\underset{z\to \mathrm{\infty }}{lim}X(z)=x(0)+x(1){\mathrm{\infty }}^{-1}+x(2){\mathrm{\infty }}^{-2}+...$

This can be written as:

$\underset{z\to \mathrm{\infty }}{lim}X(z)=x(0)$

Since x(0) is the initial value of the given signal, this is the initial value theorem of z-transform.

Since, We cannot apply the z transform, none of these is the correct answer. 

Since,$\mathrm{x}\left[\mathrm{n}\right]$ is a causal signal, we can apply initial value theorem to find relation between $\mathrm{A}$ and $\mathrm{C}$. We know, the initial value of $\mathrm{x}\left[\mathrm{n}\right]$ is $\underset{n\to 0}{lim}x\left[n\right]=1$.

Using initial value theorem, $\underset{\mathrm{n}\to 0}{lim}\mathrm{x}\left[\mathrm{n}\right]=\underset{\mathrm{z}\to \mathrm{\infty }}{lim}\mathrm{X}\left[\mathrm{z}\right]$

$\begin{array}{l}\Rightarrow \mathrm{X}\left(\mathrm{z}\right)=\underset{\mathrm{z}\to \mathrm{\infty }}{lim}\frac{\mathrm{A}{\mathrm{z}}^2+\mathrm{B}\mathrm{z}}{\mathrm{C}{\mathrm{z}}^2+\mathrm{D}\mathrm{z}+\mathrm{E}}=1\\ \Rightarrow \underset{\mathrm{z}\to \mathrm{\infty }}{lim}\frac{\mathrm{A}+\frac{\mathrm{B}}{\mathrm{z}}}{\mathrm{C}{\mathrm{z}}^2+\frac{\mathrm{D}}{\mathrm{z}}+\frac{\mathrm{E}}{{\mathrm{z}}^2}}=\frac{\mathrm{A}}{\mathrm{C}}=1\end{array}$

Now, $\frac{\mathrm{A}}{\mathrm{C}}=1$

$\Rightarrow \mathrm{A}=\mathrm{C}$

For a pole at infinity, $\begin{array}{c}Lim\\ z\to \mathrm{\infty }\end{array}\mathrm{X}\left(\mathrm{z}\right)$ should be infinity. And for a zero to exist at infinity $\begin{array}{c}Lim\\ z\to \mathrm{\infty }\end{array}\mathrm{X}\left(\mathrm{z}\right)$ should be zero. Now, from initial value theorem

$\begin{array}{c}Lim\\ z\to \mathrm{\infty }\end{array}\mathrm{X}\left(\mathrm{z}\right)=\mathrm{x}\left[0\right]=1$

Thus, neither a pole nor a zero exists at infinity. $\mathrm{x}[1]=3$ was unnecessary information.

$\mathrm{X}(\mathrm{z})$ by long division method can be written as.

$\begin{array}{l}X\left(z\right)=x\left[0\right]+x\left[1\right]z^{-1}+x\left[2\right]z^{-2}+x\left[3\right]z^{-3}\dots \dots \dots \\ \Rightarrow \begin{array}{c}Lim\\ z\to \mathrm{\infty }\end{array}{\mathrm{z}}^2\left[\mathrm{X}\left(\mathrm{z}\right)-\mathrm{x}\left[0\right]-\mathrm{x}\left(1\right).{\mathrm{z}}^{-1}\right]=\mathrm{x}\left[2\right]\end{array}$

In general,

$\Rightarrow \mathrm{x}\left[\mathrm{n}\right]=\begin{array}{c}Lim\\ z\to \mathrm{\infty }\end{array}{\mathrm{z}}^{\mathrm{n}}\left[\mathrm{X}\left(\mathrm{z}\right)-\mathrm{x}\left[0\right]-\mathrm{x}\left(1\right).{\mathrm{z}}^{-1}\dots \dots -{\mathrm{z}}^{-\left(\mathrm{n}-1\right)}\mathrm{x}\left[\mathrm{n}-1\right]\right]$

Solution: An anti causal sequence is of the form

$\begin{array}{l}x\left[n\right]=a_0+a_{-1}\delta \left[n+1\right]+a_{-2}\delta \left[n+2\right]+a_{-3}\delta \left[n+3\right]\dots \dots \dots \\ \mathrm{X}\left[\mathrm{z}\right]={\mathrm{a}}_0+{\mathrm{a}}_{-1}\mathrm{z}+{\mathrm{a}}_{-2}{\mathrm{z}}^2+{\mathrm{a}}_{-3}{\mathrm{z}}^3..\dots \dots .\dots \end{array}$

when we put $\mathrm{z}=0$ we get ${\mathrm{a}}_0$

which is the initial value

Thus, $\mathrm{x}\left[0\right]=\begin{array}{c}Lim\\ z\to 0\end{array}\mathrm{X}\left(\mathrm{z}\right)$